For a general Riemannian manifold $M$, the covariant derivative $\nabla$ admits a formal adjoint $\nabla^*$. In particular, if $S$ is a $(1, 1)$-tensor on $M$, we have $$\nabla^*S=-tr \nabla S=-\sum _i(\nabla_{e_i} S)e_i$$ where $\{e_i\}_i$ is a local orthonormal basis.

On the other hand we know that every covariant derivative $\nabla$ commutes with all contractions. That is if “ tr” denotes the trace on any pair of indices, $$tr\nabla_X A=\nabla_X tr A.$$

Q: Is it correct that formal adjoint of any trace-free $(1,1)$-tensor vanishes?



I think what you meant to ask is whether it's true that the formal adjoint of $\nabla$ applied to any trace-free $(1,1)$-tensor vanishes. (A $(1,1)$-tensor has a formal adjoint as a linear endomorphism, but that's not a differential operator.)

The answer is no.

It is true that $\operatorname{tr} \nabla_X A = \nabla_X \operatorname{tr} A$, but this doesn't have anything to do with $\nabla^*$. The formal adjoint of $\nabla$ acting on $A$ is $\nabla^* A = - \operatorname{tr} (\nabla A)$, not $-\operatorname{tr} \nabla_X A$. If you adopt the convention that the new index position introduced by $\nabla$ is the last one, then this trace is taken on the last two indices of $\nabla A$ -- one of the original indices of $A$, and the new one introduced by differentiation. In index notation, this is $$ (\nabla^* A)_i = - A^j_{i;j}. $$

On the other hand, you can also consider $\nabla(\operatorname{tr} A) = \operatorname{tr}(\nabla A)$; but in this case, the trace is on the first two indices of $\nabla A$, namely the two original indices of $A$ before the one resulting from differentiation. In index notation, $$ \big(\nabla(\operatorname{tr} A)\big)_i = A^j_{j;i} = 0. $$ As you can see, the "$\operatorname{tr}$" notation is ambiguous -- you always have to be explicit about which indices are being traced (or decorate the trace symbol with numbers, as in $\operatorname{tr}_{12}$ or $\operatorname{tr}_{23}$).

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